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# Find the angle set

Question:

Given: sin (2x + pi/3) = - 1/2. Find the set of the angle x.

Solution:

let t = 2x + pi/3, then sin (2x + pi/3) = -1/2 change to sin t = - 1/2. Because the value of sin t is negative, so the angle t is either in Quadrant III or in Quadrant IV. There is t1 in Quadrant III satisfy sin t1 = - 1/2, or there is t2 in Quadrant IV satisfy sin t2 = - 1/2.

Now we will find an acute angle that satisfy sin t = 1/2. Then t = arc sin 1/2. Because sin 30o = 1/2, so, t = arc sin 1/2 = 30o = pi/6.

So, the angle in Quadrant III is t1 = pi + pi/6 = 7pi/6. The angles that more than one cycle in Quadrant III is t = 7pi/6 + 2pi k (in which k is an integer, 2pi is one minimum positive period of y = sin t). Now back to x, 2x + pi/3 = 7pi/6 + 2pi k,

2x + pi/3 = 7pi/6 + 2pi k
2x = - pi/3 + 7pi/6 + 2pi k
2x = - 2pi/6 + 7pi/6 + 2pi k
2x = 5pi/6 + 2pi k
x = 5pi/12 + pi k

So, x = 5pi/12 + pi k, in which x is the Angles in Quadrant III and satisfy sin (2x + pi/3) = -1/2

k = 0, x1 = 5pi/12
k = 1, x1 = 5pi/12 + pi = 17pi/12
k = 2, x1 = 5pi/12 + 2pi = 29pi/12
k = 3, x1 = 5pi/12 + 3pi = 41pi/12

The angle in Quadrant IV is t2 = 2pi - pi/6 = 11pi/6. The angle that is more than one cycle in Quadrant IV is t = 11pi/6 + 2pi k (k is an integer). Now back to x. 2x + pi/3 = 11pi/6 + 2pi k

2x + pi/3 = 11pi/6 + 2pi k
2x = - pi/3 + 11pi/6 + 2pi k
2x = - 2pi/6 + 11pi/6 + 2pi k
2x = 9pi/6 + 2pi k
2x = 3pi/2 + 2pi k
x = 3pi/4 + pi k

So, x = 3pi/4 + pi k, in which x is the Angles in Quadrant IV and satisfy sin (2x + pi/3) = -1/2

k = 0, x2 = 3pi/4
k = 1, x2 = 3pi/4 + pi = 7pi/4
k = 2, x2 = 3pi/4 + 2pi = 11pi/4
k = 3, x2 = 3pi/4 + 3pi = 15pi/4

The angle set is {x| x = 5pi/12 + pi k or x = 3pi/4 + pi k} Look the figure above, the blue curve is the graph of y = sin (2x + pi/3). The orange line is y = - 1/2. The intersection points of the blue curve and the orange line is the solution of sin (2x + pi/3) = - 1/2. Watch the video for more details.